An inspector believes that 18% of cups made at a certain factory contain flaws. The factory owner claims that the true percentage is less than 18%. The inspector examines a random sample of 40 cups and finds that 3 of them contain flaws.
- (a)Stating a necessary assumption, use a binomial distribution to test the factory owner's claim at the 5% significance level.[6]
- (b)Explain why it would not be appropriate to use the Poisson approximation to the binomial distribution to carry out the test in part (a).[1]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.
- [1] (a) Assume flaws in cups are independent OR prob of containing a flaw is the same for all cups
- [1] (a) Ho: P(contains flaw) = 0.18 H1: P(contains flaw) < 0.18
- [1] (a) [P(X<=3) =] 0.82^40 + 40×0.82^39×0.18 + 40C2×0.82^38×0.18^2 + 40C3×0.82^37×0.18^3
- [1] (a) =0.0003569 + 0.0031338 +0.013414 + 0.037298
- [1] (a) 0.0542 (3 sf)
- [1] (a) 0.0542>0.05
- [1] (a) [Accept Ho] Insufficient evidence to accept factory owners claim. [Insufficient evidence that percentage less than 18%]
- [1] (b) np = 7.2 which is too large, or > 5. Or n = 40 which is not greater than 50. Or p = 0.18 which is not small or > 0.1.
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